Question

The following table shows how the specific heat at constant pressure of liquid helium changes with temperature. Note the sharp increase over this temperature range:

Temperature (K):

1.80 1.85 1.90 1.95 2.00 2.05 2.10 2.15

${\mathbf{c}}_{\mathbf{s}}\left({\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}\right)$:

2.81 3.26 3.79 4.42 5.18 6.16 7.51 9.35

Estimate how much heat it takes at constant pressure to increase the temperature of 1.00 g He(l) from 1.8 to 2.15 K. (Hint:For each temperature interval of 0.05 K,

take the average, ${\mathbf{c}}_{\mathbf{s}}$, as the sum of the values at the ends of the interval divided by 2.)

Short answer

The heat it takes is0.1635 J at constant pressure to increase the temperature of 1.00 g He(l) from 1.8 to 2.15 K.

Step-by-step solution

Given Information.

1. The volume of X is 10.00 mL.
2. The molarity of silver nitrate is 0.100 M.
3. The weight of the precipitate is 0.762 g.
4. The molarity of sodium hydroxide is 0.100 M.

The Heat.

The heat taken up or given by the system is determined by the Law of conservation of energy. It is the product of mass, specific heat, and temperature gradient.

The heat taken up and given out can be calculated using the equation;

$Q=m{C}_p\Delta T$

Here the temperature gradient is taken.

The heat for each interval can be calculated by substituting the values of $\Delta T$as 0.05 K and taking the average of $c_s$.

The amount of heat it takes at constant pressure to increase the temperature.

1) Heat when the specific heat changes from 2.81 to 3.26;

$$\begin{gathered} {\text{c}}_{\text{s}}\text{=}\frac{\text{3.26 - 2.81}}{\text{2}} \\ \text{=}\frac{\text{0.225 J}}{\text{gK}} \end{gathered}$$

The value of Q is:

$\begin{array}{c}\mathrm{Q}=1\times \frac{0.225\mathrm{J}}{\mathrm{gK}}\times 0.05\mathrm{K}\\ =0.01125\mathrm{J}\end{array}$

2) Heatwhen the specific heat changes from 3.26 to 3.79;

$\begin{array}{c}{\mathrm{c}}_{\mathrm{s}}=\frac{3.79-3.26}{2}\\ =\frac{0.265\mathrm{J}}{\mathrm{gK}}\end{array}$

The value of Q is:

$\begin{array}{c}\mathrm{Q}=1\times \frac{0.265\mathrm{J}}{\mathrm{gK}}\times 0.05\mathrm{K}\\ =0.01325\mathrm{J}\end{array}$

3) Heatwhen the specific heat changes from 3.79 to 4.42;

$\begin{array}{c}{\mathrm{c}}_{\mathrm{s}}=\frac{4.42-3.79}{2}\\ =\frac{0.315\mathrm{J}}{\mathrm{gK}}\end{array}$

The value of Q is:

$\begin{array}{c}\mathrm{Q}=1\times \frac{0.315\mathrm{J}}{\mathrm{gK}}\times 0.05\mathrm{K}\\ =0.01575\mathrm{J}\end{array}$

4) Heat when the specific heat changes from 4.42 to 5.18;

$\begin{array}{c}{\mathrm{c}}_{\mathrm{s}}=\frac{5.18-4.42}{2}\\ =\frac{0.38\mathrm{J}}{\mathrm{gK}}\end{array}$

The value of Q is;

$\begin{array}{c}\mathrm{Q}=1\times \frac{0.38\mathrm{J}}{\mathrm{gK}}\times 0.05\mathrm{K}\\ =0.019\mathrm{J}\end{array}$

5) Heat when the specific heat changes from 5.18 to 6.16.

$\begin{array}{c}{\mathrm{c}}_{\mathrm{s}}=\frac{6.16-5.82}{2}\\ =\frac{0.49\mathrm{J}}{\mathrm{gK}}\end{array}$

The value of Q is:

$\begin{array}{c}\mathrm{Q}=1\times \frac{0.49\mathrm{J}}{\mathrm{gK}}\times 0.05\mathrm{K}\\ =0.0245\mathrm{J}\end{array}$

6) Heat when the specific heat changes from 6.16 to 7.51.

$\begin{array}{c}{\mathrm{c}}_{\mathrm{s}}=\frac{7.51-6.16}{2}\\ =\frac{0.675\mathrm{J}}{\mathrm{gK}}\end{array}$

The value of Q is;

$\begin{array}{c}\mathrm{Q}=1\times \frac{0.675\mathrm{J}}{\mathrm{gK}}\times 0.05\mathrm{K}\\ =0.03375\mathrm{J}\end{array}$

7) Heat when the specific heat changes from 7.51 to 9.35.

$\begin{array}{c}{\mathrm{c}}_{\mathrm{s}}=\frac{9.35-7.51}{2}\\ =\frac{0.92\mathrm{J}}{\mathrm{gK}}\end{array}$

The value of Q is:

$\begin{array}{c}\mathrm{Q}=1\times \frac{0.92\mathrm{J}}{\mathrm{gK}}\times 0.05\mathrm{K}\\ =0.046\mathrm{J}\end{array}$

Hence, the total heat can be obtained by taking the sum of all the heats that is calculated.

$\begin{array}{c}\mathrm{Q}={\mathrm{Q}}_1+{\mathrm{Q}}_2+{\mathrm{Q}}_3+{\mathrm{Q}}_4+{\mathrm{Q}}_5+{\mathrm{Q}}_6+{\mathrm{Q}}_7\\ =0.01125\mathrm{J}+0.01325\mathrm{J}+0.01575\mathrm{J}+0.019\mathrm{J}+0.0245\mathrm{J}+0.03375\mathrm{J}+0.046\mathrm{J}\\ =0.1635\mathrm{J}\end{array}$

Hence, the heat it takes is 0.1635 J.